{"id":218002,"date":"2025-05-23T05:44:17","date_gmt":"2025-05-23T05:44:17","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=218002"},"modified":"2025-05-23T05:44:19","modified_gmt":"2025-05-23T05:44:19","slug":"enough-of-a-monoprotic-acid-is-dissolved-in-water-to-produce-a-0-0192-m-solution-30","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/23\/enough-of-a-monoprotic-acid-is-dissolved-in-water-to-produce-a-0-0192-m-solution-30\/","title":{"rendered":"Enough of a monoprotic acid is dissolved in water to produce a 0.0192 M solution."},"content":{"rendered":"\n<p class=\"wp-block-paragraph\">Enough of a monoprotic acid is dissolved in water to produce a 0.0192 M solution. The pH of the resulting solution is 2.53. Calculate the Ka for the acid.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-1-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">To find the <strong>Ka<\/strong> of a monoprotic acid from its concentration and the pH of its solution, we follow these steps:<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Given:<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Concentration of acid, <strong>[HA] = 0.0192 M<\/strong><\/li>\n\n\n\n<li>pH = <strong>2.53<\/strong><\/li>\n\n\n\n<li>The acid is <strong>monoprotic<\/strong>, meaning it donates one proton per molecule.<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Step 1: Calculate [H\u207a] from pH<\/strong><\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">We use the formula:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>[\\text{H}^+] = 10^{-\\text{pH}} = 10^{-2.53} \\approx 2.95 \\times 10^{-3} \\, \\text{M}<br>$$<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Step 2: Set up the equilibrium expression<\/strong><\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">The dissociation of a monoprotic acid is:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>\\text{HA} \\rightleftharpoons \\text{H}^+ + \\text{A}^-<br>$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Initial concentrations:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>[HA] = 0.0192 M<\/li>\n\n\n\n<li>[H\u207a] = 0<\/li>\n\n\n\n<li>[A\u207b] = 0<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">Change in concentration:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>[HA] decreases by $x$<\/li>\n\n\n\n<li>[H\u207a] and [A\u207b] increase by $x$<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">At equilibrium:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>[HA] = 0.0192 &#8211; x<\/li>\n\n\n\n<li>[H\u207a] = x<\/li>\n\n\n\n<li>[A\u207b] = x<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">Since we calculated $[H^+] = 2.95 \\times 10^{-3}$, we let:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>x = 2.95 \\times 10^{-3}<br>$$<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Step 3: Write the Ka expression<\/strong><\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>K_a = \\frac{[\\text{H}^+][\\text{A}^-]}{[\\text{HA}]} = \\frac{x^2}{0.0192 &#8211; x}<br>$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Plug in the values:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>K_a = \\frac{(2.95 \\times 10^{-3})^2}{0.0192 &#8211; 2.95 \\times 10^{-3}} = \\frac{8.70 \\times 10^{-6}}{0.01625}<br>$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>K_a \\approx 5.35 \\times 10^{-4}<br>$$<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">\u2705 <strong>Final Answer:<\/strong><\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>\\boxed{K_a \\approx 5.35 \\times 10^{-4}}<br>$$<\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">The acid dissociation constant, <strong>Ka<\/strong>, is a quantitative measure of the strength of a weak acid in solution. It is derived from the equilibrium concentrations of the reactants and products in the dissociation of an acid in water. Since this problem involves a <strong>monoprotic acid<\/strong>, we assume it releases <strong>one proton (H\u207a)<\/strong> per molecule of acid (HA) that dissociates.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">We are given the molarity of the acid solution and the pH. The pH is a measure of hydrogen ion concentration and is calculated using the formula:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>\\text{pH} = -\\log[\\text{H}^+]<br>$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Reversing this formula allows us to determine the actual hydrogen ion concentration, which represents the amount of acid that has dissociated. With this, we assume that the concentration of <strong>A\u207b<\/strong> (the conjugate base) is equal to [H\u207a] because of the 1:1 dissociation ratio of monoprotic acids.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Using this information, we apply the <strong>Ka expression<\/strong>, which is:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>K_a = \\frac{[\\text{H}^+][\\text{A}^-]}{[\\text{HA}]}<br>$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Here, the initial concentration of the acid is reduced by the amount that dissociates ($x$), and this amount is equal to the concentrations of H\u207a and A\u207b at equilibrium. By plugging in these values, we derive the numerical value for Ka.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">The final result of $\\boxed{5.35 \\times 10^{-4}}$ tells us the extent to which the acid dissociates in water. This Ka value indicates a <strong>moderately weak acid<\/strong>, which partially dissociates, consistent with the given pH of 2.53. This method is commonly used in acid-base equilibrium calculations in chemistry.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/05\/image-114.png\" alt=\"\" class=\"wp-image-218003\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Enough of a monoprotic acid is dissolved in water to produce a 0.0192 M solution. The pH of the resulting solution is 2.53. Calculate the Ka for the acid. The Correct Answer and Explanation is: To find the Ka of a monoprotic acid from its concentration and the pH of its solution, we follow these [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-218002","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/218002","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=218002"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/218002\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=218002"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=218002"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=218002"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}